Question

Let, $\omega$ be a complex cube root of unity with $\omega \ne 1$. A fair die is thrown three times. If $r_1,r_2$ and $r_3$ are the numbers obtained on the die, then the probability that ${\omega }^{r_1}+{\omega }^{r_2}+{\omega }^{r_3}=0$ is

• A. $\frac{1}{18}$
• B. $\frac{1}{9}$
• C. $\frac{2}{9}$
• D. $\frac{1}{36}$

Answer 1

The correct option is C $\frac{2}{9}$

A dice is thrown thrice $n\left(s\right)=6\times 6\times 6$

Favorable events, $W^{r_1}+W^{r_2}+W^{r_3}=0$

$(r_1,r_2,r_3)$ are ordered 3 triples which can take values.

$(1,2,3)(1,5,3)(4,2,3)(4,5,3)$

$(1,2,6)(1,5,6)(4,2,6)(4,5,6)$

i.e, 8 ordered pairs and each can be arranged in 3! ways.

$3!=6$

$n\left(E\right)=8\times 6$

$P\left(E\right)=\frac{8\times 6}{6\times 6\times 6}=\frac{2}{9}$